What is the solution for (dy)/(dx)=(e^{2x-y})/(e^{x+y)} ?

The solution for (dy)/(dx)=(e^{2x-y})/(e^{x+y)} is y=(ln(2e^x+c_{1}))/2

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(dy)/(dx)=(e^{2x-y})/(e^{x+y)}

\frac{d y}{d x} = \frac{e ^{2 x - y}}{e ^{x + y}}

y = \frac{ln ( 2 e ^{x} + c _{1} )}{2}

Solution steps

\frac{d y}{d x} = \frac{e ^{2 x - y}}{e ^{x + y}}

Solve separable ODE:

y = \frac{ln ( 2 e ^{x} + c _{1} )}{2}

y = \frac{ln ( 2 e ^{x} + c _{1} )}{2}

\int \frac{1}{1 - 2 x} - cos (\frac{x}{3}) d x

\frac{\partial}{\partial x} (yz e^{x})

y^{^{''}} - y^{^{'}} - 2 y = - 4 t + 6 t^{2}

\int x e^{6} d x

What is the solution for (dy)/(dx)=(e^{2x-y})/(e^{x+y)} ?
The solution for (dy)/(dx)=(e^{2x-y})/(e^{x+y)} is y=(ln(2e^x+c_{1}))/2